In the first system, the slopes are different but they have one intersection. In the second system they have the same slopes but no intersection. In the third system they again have the same slope but no intersection.
2.2:
There are three different methods for solving a system of a equations. The first one is called substitution. This is when you look at a system of equations and rewrite one of them into slope-intercept form so you have the value of either x or y. Then you plug that value into the other system equation and solve for the other variable. The second method to finding a solution to a system of equations is called elimination. To use this method, you first have to get one of the variables by itself. So if you have 2y and y, you would need to get y by itself by putting a negative 2 in front of the y because 2+-2=0. Then you distribute that co-efficient to the other numbers in the equation and add both systems together. That gives you the solution to one of the variables but then you have to plug that information into the other equation to solve for the other variable. The third method you can use is graphing. To do this you can use substitution to find one of the points and plot it on the graph. To find another point, plug zero into the x or y variable to find the x or y-intercept. Then plot that point on the graph and connect your two points with a line. I would use substitution when one of the equations already looks like it is close to slope-intercept form. That way, it is easier to solve and there isn't much rewriting to do. Elimination can be used when you have one variable by itself already. I would graph the systems if the neither of the variables in the equations were alone and I couldn't eliminate easily. For elimination, I have to remember to distribute the co-efficient to the other numbers in the equation. If I don't do that then my whole answer will be off so it is important to remember that. The three methods to solving a system of equations are substitution, elimination, and graphing.
2.3:
y = 75x + 1050
The y-intercept of the blue linear function is 1000. The y-intercept of the red linear function is 1050. This means that this is how much money both bank accounts began with.
The slope of the blue linear function is 100.The slope of the red linear function is 75.This means that at the end of each year, the people deposited that much extra into their bank accounts.
The bank account for the blue linear function is better. Yes, they start off with $50 dollars less than the red linear function bank account, but they deposit $100 into their bank account at the end of the year while the red linear function deposits $75 at the end of the year. For the first three years, the red linear function bank account had more money than the blue bank account. On the third year, they both had $1200 in their bank account and then the blue linear function started to have more money in their bank account.
I would choose to open up the blue linear function bank account to save up for college because the money accumulates faster in that bank account. The blue bank account gets $25 more than the red bank account at the end of the year.
I would also choose to open up the blue linear function bank account for my own child. At first, the red bank account has more money but then the amount of money in the blue bank account increases faster than the red linear function bank account.
In the first system, the slopes are different but they have one intersection. In the second system they have the same slopes but no intersection. In the third system they again have the same slope but no intersection.
2.2:
There are three different methods for solving a system of a equations. The first one is called substitution. This is when you look at a system of equations and rewrite one of them into slope-intercept form so you have the value of either x or y. Then you plug that value into the other system equation and solve for the other variable. The second method to finding a solution to a system of equations is called elimination. To use this method, you first have to get one of the variables by itself. So if you have 2y and y, you would need to get y by itself by putting a negative 2 in front of the y because 2+-2=0. Then you distribute that co-efficient to the other numbers in the equation and add both systems together. That gives you the solution to one of the variables but then you have to plug that information into the other equation to solve for the other variable. The third method you can use is graphing. To do this you can use substitution to find one of the points and plot it on the graph. To find another point, plug zero into the x or y variable to find the x or y-intercept. Then plot that point on the graph and connect your two points with a line. I would use substitution when one of the equations already looks like it is close to slope-intercept form. That way, it is easier to solve and there isn't much rewriting to do. Elimination can be used when you have one variable by itself already. I would graph the systems if the neither of the variables in the equations were alone and I couldn't eliminate easily. For elimination, I have to remember to distribute the co-efficient to the other numbers in the equation. If I don't do that then my whole answer will be off so it is important to remember that. The three methods to solving a system of equations are substitution, elimination, and graphing.
2.3:
y = 75x + 1050
The y-intercept of the blue linear function is 1000. The y-intercept of the red linear function is 1050. This means that this is how much money both bank accounts began with.
The slope of the blue linear function is 100. The slope of the red linear function is 75. This means that at the end of each year, the people deposited that much extra into their bank accounts.
The bank account for the blue linear function is better. Yes, they start off with $50 dollars less than the red linear function bank account, but they deposit $100 into their bank account at the end of the year while the red linear function deposits $75 at the end of the year. For the first three years, the red linear function bank account had more money than the blue bank account. On the third year, they both had $1200 in their bank account and then the blue linear function started to have more money in their bank account.
I would choose to open up the blue linear function bank account to save up for college because the money accumulates faster in that bank account. The blue bank account gets $25 more than the red bank account at the end of the year.
I would also choose to open up the blue linear function bank account for my own child. At first, the red bank account has more money but then the amount of money in the blue bank account increases faster than the red linear function bank account.